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Sun Oct 14, 2012, 03:35 PM

Very simple geometry question...

Since I have entered into my 50th year I have forgotten most if not all of my basic geometry. (I was never a wiz at math). However, I know the caveman way to figure out the following problem but I wish to know the elegant mathematical solution to this problem. Thanks!

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Reply Very simple geometry question... (Original post)
Javaman Oct 2012 OP
HereSince1628 Oct 2012 #1
unc70 Oct 2012 #2
BootinUp Oct 2012 #3
immoderate Oct 2012 #4
dipsydoodle Oct 2012 #5
Jim__ Oct 2012 #6
randr Oct 2012 #7
dipsydoodle Oct 2012 #9
randr Oct 2012 #11
Posteritatis Oct 2012 #12
Tuesday Afternoon Oct 2012 #13
Chemisse Oct 2012 #14
TheMadMonk Oct 2012 #15
Chemisse Oct 2012 #16
TheMadMonk Oct 2012 #20
opiate69 Oct 2012 #17
TheMadMonk Oct 2012 #19
montanto Oct 2012 #18
TheMadMonk Oct 2012 #21
dimbear Oct 2012 #22
erinlough Oct 2012 #8
intaglio Oct 2012 #10
Confusious Oct 2012 #23
rerevisionist Oct 2012 #24

Response to Javaman (Original post)

Sun Oct 14, 2012, 03:41 PM

1. With a compass and straight edge or by the numbers?

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:41 PM

2. 180-45 degrees. If I understand your drawing

Think of the line as bisecting a circle. Half of 360 degrees means 180 degrees on either side. Since the section dividing that side has 45 degrees in one direction, it must have 180-45 = 135 degrees in the other direction.

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:41 PM

3. 180 - 45?

Or is that the caveman approach.

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:43 PM

4. You are not really pointing to an angle, but I think you mean 22.5 degrees.

Divide by 2?

--imm

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:44 PM

5. Not sure which angle you mean

but each side of the dotted line is 22.5 degrees.

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:55 PM

6. Are you pointing at the exterior (obtuse) angle between the intersection of the 2 beams?

Given that the sides of each beam are parallel, that exterior angle is 1350 - the same as the angle between the implicit continuation of the straight line (the side of the beam that is right above it) and the beam that is intersecting it (the beam that makes the 450 acute angle with the line).

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Response to Javaman (Original post)

Sun Oct 14, 2012, 03:55 PM

7. I just came across this

This was sent to me and it gave me a laugh. No known author, one of those chain emails and it obviously never happened. Still it may pose possible means of solving your dilemma.

"Thinking Unconventionally
A Letter from a College Professor

Some time ago I received a call from a colleague, who asked if I would be the referee on the grading of an examination question. He was about to give a student a zero for his answer to a physics question, while the student claimed he should receive a perfect score and would if the system were not set up against the student.

The instructor and the student agreed to an impartial arbiter, and I was selected. I went to my colleague's office and read the examination question: "Show how it is possible to determine the height of a tall building with the aid of a barometer."

The student had answered: "Take the barometer to the top of the building, attach a long rope to it, lower it to the street, and then bring it up, measuring the length of the rope. The length of the rope is the height of the building."

I pointed out that the student really had a strong case for full credit since he had really answered the question completely and correctly. On the other hand, if full credit were given, it could well contribute to a high grade in his physics course. A high grade is supposed to certify competence in physics, but the answer did not confirm this. I suggested that the student have another try at answering the question. I was not surprised that my colleague agreed, but I was surprised when the student did.

I gave the student six minutes to answer the question with the warning that the answer should show some knowledge of physics. At the end of five minutes, he had not written anything. I asked if he wished to give up, but he said no. He had many answers to this problem; he was just thinking of the best one. I excused myself for interrupting him and asked him to please go on. In the next minute, he dashed off his answer which read:

"Take the barometer to the top of the building and lean over the edge of the roof. Drop the barometer, timing its fall with a stopwatch. Then, using the formula x=3D0.5*a*t^2, calculate the height of the building."

At this point, I asked my colleague if he would give up. He conceded, and gave the student almost full credit. In leaving my colleague's office, I recalled that the student had said that he had other answers to the problem, so I asked him what they were.

"Well," said the student. "there are many ways of getting the height of a tall building with the aid of a barometer. For example, you could take the barometer out on a sunny day and measure the height of the barometer, the length of its shadow, and the length of the shadow of the building, and by the use of simple proportion, determine the height of the building."

"Fine," I said, "and others?"

"Yes," said the student." There is a very basic measurement method you will like. In this method, you take the barometer and begin to walk up the stairs. As you climb the stairs, you mark off the length of the barometer along the wall. You then count the number of marks, and this will give you the height of the building in barometer units.

"A very direct method."

"Of course. If you want a more sophisticated method, you can tie the barometer to the end of a string, swing it as a pendulum, and determine the value of g at the street level and at the top of the building. From the difference between the two values of g, the height of the building, in principle, can be calculated."

"On this same tack, you could take the barometer to the top of the building, attach a long rope to it, lower it to just above the street, and then swing it as a pendulum. You could then calculate the height of the building by the period of the precession".

"Finally," he concluded, "there are many other ways of solving the problem. Probably the best," he said, "is to take the barometer to the basement and knock on the superintendent's door. When the superintendent answers, you speak to him as follows: 'Mr. Superintendent, here is a fine barometer. If you will tell me the height of the building, I will give you this barometer.'"

At this point, I asked the student if he really did know the conventional answer to this question. He admitted that he did, but said that he was fed up with high school and college instructors trying to "teach him to think"."

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Response to randr (Reply #7)

Sun Oct 14, 2012, 04:40 PM

9. That is without a doubt

one of the best replies I have ever read here on DU.

You are a star.

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Response to dipsydoodle (Reply #9)

Sun Oct 14, 2012, 05:31 PM

11. Thank you kind sir

I will pass the star to my wife who passed it to me.

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Response to randr (Reply #7)

Sun Oct 14, 2012, 06:20 PM

12. I never get tired of the superintendent option. (nt)

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Response to randr (Reply #7)

Sun Oct 14, 2012, 07:16 PM

13. awesome

student

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Response to randr (Reply #7)

Sun Oct 14, 2012, 08:56 PM

14. Nice story.

And some very good ideas on how to solve it. So what is the conventional answer to this?

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Response to Chemisse (Reply #14)

Tue Oct 16, 2012, 11:49 PM

15. Difference in air pressure between top and bottom of building.

 

As for the story, I'd Snopes it. I recall a very similar story told as a lesson in latteral thinking from my high school days.

Here's another from a similar source.

From my starting point I travel 1 Km North, 1 East and 1 South returning to my starting point. Where am I? And are you absolutely certain?

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Response to TheMadMonk (Reply #15)

Thu Oct 18, 2012, 12:30 PM

16. Thanks.

Hmm. So are you 1 km east of where you started? Or does the lack of units after the second two numbers mean you may have traveled 1 mile or 1 cm, etc?

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Response to Chemisse (Reply #16)

Thu Oct 18, 2012, 08:30 PM

20. Lack of subsequent units implies the same unit is used throuout.

 

And the starting point and ending point are the same.

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Response to TheMadMonk (Reply #15)

Thu Oct 18, 2012, 12:39 PM

17. The South Pole. But no, I`m not absolutely certain.. close to it, though.

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Response to opiate69 (Reply #17)

Thu Oct 18, 2012, 08:27 PM

19. South pole will get you a passing grade.

 

For bonus points: Anywhere on a circle 2318.31m (1000/pi + 1000 x 2) in diameter centred on the NORTH pole. Plus a concentric series of smaller circles where a whole number of circuts (after traveling the 1000m North) adds up to 1000m.

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Response to TheMadMonk (Reply #15)

Thu Oct 18, 2012, 04:43 PM

18. in Antarctica.

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Response to montanto (Reply #18)

Thu Oct 18, 2012, 08:33 PM

21. See reply to 17 above. /nt

 

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Response to randr (Reply #7)

Fri Oct 19, 2012, 08:18 PM

22. Hmmm. I could see how the precession would give you your latitude. I don't see

how it could give you the height of the building. And that's measuring the extent of the precession, not the period, which is inevitably an astronomical day.

Willing to be set straight.

I think he should have said simply "the period of the pendulum."

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Response to Javaman (Original post)

Sun Oct 14, 2012, 04:22 PM

8. If the two rectangles are equal in width then the angle

Is 180-45 to give you the measure of the outside obtuse angle.

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Response to Javaman (Original post)

Sun Oct 14, 2012, 05:18 PM

10. From first principles

The upper lines intersect each other at 45 degrees.

That gives you 2 of the angles at the intersection, 45 and 45, the remaining angles at the intersection total 270 degrees (45 + 45 + 270 = 360). Those 2 angles are identical so each is 270/2 = 135 degrees.

The dotted line bisects the 2 angles it is drawn between so the small right angled triangle has angles of
135/2 = 67.5 degrees
and 45/2 = 22.5 degrees at the base.

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Response to Javaman (Original post)

Mon Oct 22, 2012, 05:37 PM

23. Similar triangles.

The upper rectangle bottom is 90 degrees from the 45 angles starting line, and the lower rectangle side is 45, so the lower rectangles top makes a 45 degree with the lower rectangle bottom.

Cut that in half, 22.5 degrees.

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Response to Javaman (Original post)

Mon Oct 29, 2012, 02:10 PM

24. (90 - 45) /2 degrees = 22.5

Assuming you're being Euclidean.

And assuming your wood blocks are all rectangular.

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